Prove that: \[\sqrt{\frac{1 - \sin\theta}{1 + \sin\theta}} = \sec\theta - \tan\theta\]

\[sec\theta = \frac{1}{\cos\theta}\] \[tan\theta = \frac{\sin\theta}{\cos\theta}\] \[sin^2\theta + \cos^2\theta = 1\] \[frac{1}{\cos^2\theta} = \tan^2\theta + 1\] Starting with the left-hand side: \[\sqrt{\frac{1 - \sin\theta}{1 + \sin\theta}}\] \[= \sqrt{\frac{(1 - \sin\theta)^2}{(1 - \sin^2\theta)}}\] \[= \sqrt{\frac{(1 - \sin\theta)^2}{\cos^2\theta}}\] \[= \frac{1 - \sin\theta}{\cos\theta}\] Now we can substitute the identity for \[\cos\theta:\] \[= \frac{1 - \sin\theta}{\frac{1}{\sec\theta}}\] \[= \sec\theta(1 - \sin\theta)\] \[= \sec\theta - \sec\theta\sin\theta\] \[= \sec\theta - \frac{\sin\theta}{\cos\theta}\] \[= \sec\theta - \tan\theta\] Therefore, we have shown that \[ \sqrt{\frac{1 - \sin\theta}{1 + \sin\theta}} = \sec\theta - \tan\theta.\]